Difference between revisions of "2022 AMC 10A Problems/Problem 19"
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As in solution 1, we express the LHS as a sum under one common denominator. We note that <cmath>\frac{1}{1} + \frac{1}{2} + \dots + \frac{1}{17} = \frac{\frac{17!}{1}}{17!} + \frac{\frac{17!}{2}}{17!} + \frac{\frac{17!}{3}}{17!} + \dots + \frac{\frac{17!}{17}}{17!}</cmath> | As in solution 1, we express the LHS as a sum under one common denominator. We note that <cmath>\frac{1}{1} + \frac{1}{2} + \dots + \frac{1}{17} = \frac{\frac{17!}{1}}{17!} + \frac{\frac{17!}{2}}{17!} + \frac{\frac{17!}{3}}{17!} + \dots + \frac{\frac{17!}{17}}{17!}</cmath> | ||
− | Now, we have <math>h = L_{17}\left(\frac{\frac{17!}{1} + \frac{17!}{2} + \frac{17!}{3} + \dots + \frac{17!}{17}}{17!}\right)</math>. We'd like to find <math>h \pmod{17},</math> so we can evaluate our expression <math>\pmod{17}.</math> Since <math>\frac{\frac{17!}{1}}{17!}, \frac{\frac{17!}{2}}{17!}, \dots, \frac{\frac{17!}{16}}{17!}</math> are all integers, and since <math>L_{17}</math> is a multiple of <math>17,</math> multiplying each of those terms and adding them will get a multiple of <math>17.</math> Mod 17, that result is <math>0.</math> Thus, we only need to consider <math>L_{17}\cdot \frac{\frac{17!}{17}}{17!} = \frac{L_{17}}{17} \pmod{17}.</math> Proceed with solution <math>1</math> to get <math>\boxed{\textbf{(C) 5}</math>. | + | Now, we have <math>h = L_{17}\left(\frac{\frac{17!}{1} + \frac{17!}{2} + \frac{17!}{3} + \dots + \frac{17!}{17}}{17!}\right)</math>. We'd like to find <math>h \pmod{17},</math> so we can evaluate our expression <math>\pmod{17}.</math> Since <math>\frac{\frac{17!}{1}}{17!}, \frac{\frac{17!}{2}}{17!}, \dots, \frac{\frac{17!}{16}}{17!}</math> are all integers, and since <math>L_{17}</math> is a multiple of <math>17,</math> multiplying each of those terms and adding them will get a multiple of <math>17.</math> Mod 17, that result is <math>0.</math> Thus, we only need to consider <math>L_{17}\cdot \frac{\frac{17!}{17}}{17!} = \frac{L_{17}}{17} \pmod{17}.</math> Proceed with solution <math>1</math> to get <math>\boxed{\textbf{(C) 5}}</math>. |
~sirswagger21 | ~sirswagger21 |
Revision as of 08:46, 17 November 2022
Problem
Define as the least common multiple of all the integers from to inclusive. There is a unique integer such that What is the remainder when is divided by ?
Solution
Notice that contains the highest power of every prime below . Thus, .
When writing the sum under a common fraction, we multiply the denominators by divided by each denominator. However, since is a multiple of , all terms will be a multiple of until we divide out , and the only term that will do this is . Thus, the remainder of all other terms when divided by will be , so the problem is essentially asking us what the remainder of divided by is. This is equivalent to finding the remainder of divided by .
We use modular arithmetic to simplify our answer:
This is congruent to
Evaluating, we get: Therefore the remainder is .
~KingRavi
~mathboy282
~Scarletsyc
Solution 2
As in solution 1, we express the LHS as a sum under one common denominator. We note that
Now, we have . We'd like to find so we can evaluate our expression Since are all integers, and since is a multiple of multiplying each of those terms and adding them will get a multiple of Mod 17, that result is Thus, we only need to consider Proceed with solution to get .
~sirswagger21
Video Solution By ThePuzzlr
~ MathIsChess
See Also
2022 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 18 |
Followed by Problem 20 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.